Integral Calculus: Basic Integration of cotxln(sinx)

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Homework Help Overview

The discussion revolves around evaluating the integral of the function cot(x) multiplied by ln(sin(x)). Participants are exploring basic integration techniques from calculus, specifically avoiding advanced methods such as integration by parts or trigonometric substitution.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of substitution, specifically setting u = sin(x), and express uncertainty about how to incorporate cot(x) into their attempts. There is also a question regarding the notation of the integral.

Discussion Status

Some participants have made progress by transforming the integral into a form involving ln(u) and have derived an expression for the integral. However, there is no explicit consensus on the correctness of the approach, and one participant notes the importance of including the constant of integration.

Contextual Notes

Participants are constrained by homework rules that limit the techniques they can use for solving the integral, which may impact their approaches and reasoning.

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Homework Statement


evaluate the integral.

Homework Equations


[itex]\int (cotx)[ln(sinx)] dx[/itex]

how would i do this using basic integration techniques from calculus 1? i am not allowed to use, int by parts, partial fractions, trig sub, etc..

The Attempt at a Solution


i tried doing a u=sinx but don't know how to get the cotx.
 
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whatlifeforme said:

Homework Statement


evaluate the integral.

Homework Equations


[itex]\int (cotx)[ln(sinx)][/itex]

how would i do this using basic integration techniques from calculus 1? i am not allowed to use, int by parts, partial fractions, trig sub, etc..

The Attempt at a Solution


i tried doing a u=sinx but don't know how to get the cotx.
Where's the "dx" ?

[itex]\displaystyle \int \cot(x)\ln(\sin(x))\,dx[/itex]

This may help. [itex]\displaystyle \ \cot(x)=\frac{\cos(x)}{\sin(x)}[/itex]
 
thanks.

so i have:

u=sinx; du=cosx

[itex]\int \frac{ln(u)}{u} du[/itex]

v=lnu; dv=(1/u) du

[itex]\int v dv[/itex] = [itex]\frac{v^2}{2}[/itex] = [itex]\frac{(lnu)^2}{2}[/itex]

= [itex]\frac{[ln(sinx)]^2}{2}[/itex]
 
whatlifeforme said:
thanks.

so i have:

u=sinx; du=cosx

[itex]\int \frac{ln(u)}{u} du[/itex]

v=lnu; dv=(1/u) du

[itex]\int v dv[/itex] = [itex]\frac{v^2}{2}[/itex] = [itex]\frac{(lnu)^2}{2}[/itex]

= [itex]\frac{[ln(sinx)]^2}{2}[/itex]

Looks good to me, other than the +C.
 

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